Question

In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor. Hydrogen + Oxygen `->` water. If you burn 58.1 g of hydrogen and produce 519 g of water how much oxygen reacted?

Answer 1

`H_2(g) + 1/2O_2(g) rarr H_2O(l)`

Clearly, approx. `500` `g` of dioxygen reacted.

Explanation:

The stoichiometrically balanced equation tells us unequivocally that `1` `mol` dihydrogen reacts with `0.50` `mol` dioxygen to give `1` `mol` water.

`"Moles of dihydrogen"` `=` `(58.1*g)/(2.02*g*mol)` `=` `28.7*mol` `H_2`

`"Moles of water"` `=` `(519.0*g)/(18.02*g*mol)` `=` `28.7*mol` `H_2O`

The water is stoichiometric with respect to the amount of dihydrogen. (What do I mean by this?) There was thus a stoichiometric quantity of dioxygen gas, i.e. `14.4*mol` `O_2`

`"Mass of dioxygen"` `=` `14.4*cancel(mol)xx32.00*g*cancel(mol^-1)` `=` `??g`